How Fast Does a Ball Roll Down an Inclined Plane?

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Homework Help Overview

The discussion revolves around a physics problem involving a ball rolling down an inclined plane. The ball has a specified mass, radius, and moment of inertia, and the problem seeks to determine its speed after rolling a certain distance down the incline. The angle of the incline is also provided.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation of energy principles and question the forces acting on the ball. There are inquiries about the moment of inertia used in the calculations and whether the original poster's formula is correct.

Discussion Status

There is an ongoing exploration of the original poster's approach, with some participants suggesting that there may be a misunderstanding regarding the forces involved in rolling versus sliding. Multiple interpretations of the problem are being considered, and some participants have noted potential errors in calculation without reaching a consensus on the exact issue.

Contextual Notes

Participants are navigating the complexities of rolling motion and the forces at play, while also addressing the implications of the given moment of inertia. There is an acknowledgment of possible calculator errors in the original poster's attempts.

sheri1987
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Homework Statement



A ball of mass 2.30 kg and radius 0.142 m is released from rest on a plane inclined at an angle q = 42.0° with respect to the horizontal. How fast is the ball moving (in m/s) after it has rolled a distance d=1.35 m? Assume that the ball rolls without slipping, and that its moment of inertia about its center of mass is 1.70E-2 kg·m2.


Homework Equations



.5mv^2+.5I(v^2/R^2) = mgdsin(angle)

The Attempt at a Solution



I plugged in the numbers to equation above and got 3.2 m/s by solving for v...am I doing something wrong, because it is not the right answer?!?
 
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sheri1987 said:
.5mv^2+.5I(v^2/R^2) = mgdsin(angle)
Your error is assuming that the only force acting on the ball parallel to the plane is the component of its weight. Not so. (What other force must act so that the ball rolls instead of slides?)

Edit: I was wrong--your equation is fine.
 
Last edited:
Doc Al said:
Your error is assuming that the only force acting on the ball parallel to the plane is the component of its weight. Not so. (What other force must act so that the ball rolls instead of slides?)

I don't see the error in the OP's formula. The force you are thinking about does not do any work.

to the OP: What equation did you use for the moment of inertia? The one for a full sphere?

EDIT: never mind my last question, I had not noticed that the moment of inertia itself was given. Then it must simply be a mistake of plugging in the calculator.
 
Last edited:
I am not sure what other force, that is probably my problem
 
Doc Al said:
Your error is assuming that the only force acting on the ball parallel to the plane is the component of its weight. Not so. (What other force must act so that the ball rolls instead of slides?)

The OP is just using a conservation of energy equation. How can that be wrong? I put the numbers in and got a similar number but somewhat larger. I think it's just some kind of calculator slip up.
 
nrqed said:
I don't see the error in the OP's formula. The force you are thinking about does not do any work.

to the OP: What equation did you use for the moment of inertia? The one for a full sphere?

It looks like the moment of inertia was given in the problem statement.
 
nrqed said:
I don't see the error in the OP's formula. The force you are thinking about does not do any work.
You're right--my bad!

to the OP: What equation did you use for the moment of inertia? The one for a full sphere?
No equation needed--the moment of inertia is given.

Dick said:
The OP is just using a conservation of energy equation. How can that be wrong? I put the numbers in and got a similar number but somewhat larger. I think it's just some kind of calculator slip up.
Yep, you're right. :redface:
 
Dick said:
It looks like the moment of inertia was given in the problem statement.

Yes, thank you. I edited my post when I noticed that.
 

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